College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.8 - Page 319: 54

Answer

Center: $(-4,-2)$ $;$ Radius: $2$

Work Step by Step

$x^{2}+y^{2}+8x+4y+16=0$ Take $16$ to the right side: $x^{2}+y^{2}+8x+4y=-16$ Group the terms with $x$ together and the terms with $y$ together: $(x^{2}+8x)+(y^{2}+4y)=-16$ Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside each parentheses and also on the right side of the equation. For the first group, $b=8$ and for the second group, $b=4$ $\Big[x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}\Big]+\Big[y^{2}+4y+\Big(\dfrac{4}{2}\Big)^{2}\Big]=-16+\Big(\dfrac{8}{2}\Big)^{2}+\Big(\dfrac{4}{2}\Big)^{2}$ $(x^{2}+8x+16)+(y^{2}+4y+4)=-16+16+4$ $(x^{2}+8x+16)+(y^{2}+4y+4)=4$ Factor the expressions inside each parentheses, which are perfect square trinomials: $(x+4)^{2}+(y+2)^{2}=4$ This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$ It can be seen from the obtained equation that $(h,k)=(-4,-2)$, so the center of the circle is the point $(-4,-2)$ The radius is: $r^{2}=4$ $\sqrt{r^{2}}=\sqrt{4}$ $r=2$ The graph is:
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